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I am Lyosha [343]
3 years ago
12

a circus performer launches himself from a springboard with an initial velocity of 21 m/s at an angle of 75 toward a platform ha

nging 20 m above the ground, a horizontal distance of 15 m away from the springboard. does he land on the platform or fall back down to the ground?

Physics
1 answer:
kherson [118]3 years ago
8 0

Answer:

The circus performer falls back down to the ground

Explanation:

The question parameters are;

The initial velocity of the circus performer = 21 m/s

The angle in which the performer launches himself = 75° towards the platform

The height of the platform above the ground = 20 m

The horizontal distance of the platform from the springboard = 15 m

The vertical motion of the circus performer is given by the following projectile motion relation;

y = y₀ + v₀·sinθ₀t-1/2·g·t²

Where;

y = Height reached by the circus performer

y₀ = Initial height of the the circus performer (the springboard) = 0 m

v₀ = Initial velocity of the the circus performer = 21 m/s

θ₀ = The angle with which the circus performer launches himself = 75°

t = The time of ,light of the circus performer

g = The acceleration due to gravity

Therefore, when the height is 20 m, we have;

20 = 21*sin(75)*t - 1/2*9.81*t²

Which gives;

21*sin(75)*t - 1/2*9.81*t² - 20 = 0

Factorizing using a graphing calculator, gives;

t = 1.623 or t = 2.513

Therefore, the circus performer passes the 20 m mark twice, in his motion, where the first time is when he is on his way up while the second time is when he is on his way down

The horizontal motion of the circus performer is given by the following projectile motion relation;

x = x₀ + v₀*cos(θ₀)* t

Where;

x₀ = The initial position of the circus performer in relation to the final position = 0

Plugging in the value of t when y = 20, we get;

x = 21×cos(75)×1.623 = 8.82 m, which is less than the 15 m platform distance from the spring board

Checking the other time value, we have;

x = 21×cos(75)×2.513 = 13.66 m which is also less than the 15 m platform distance from the spring board

Therefore, the circus performer misses the platform and falls back down to the ground.

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4 years ago
Young's experiment is performed with light of wavelength 502 nm from excited helium atoms. Fringes are measured carefully on a s
jok3333 [9.3K]

Answer:

0.88 mm

Explanation:

Given

Wavelength of the atoms, λ = 502 nm = 502*10^-9 m

Radius of the screen away from the double slit, r = 1 m

We know that Y(20) = 11.4*10^-3 m

d = (20 * R * λ) / Y(20)

d = (20 * 1 * 502*10^-9)/11.4*10^-3

d = 1*10^-5 / 11.4*10^-3

d = 0.88 mm

Therefore, the distance of separation between the two slits is 0.88 mm

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4 years ago
Which theory (or theories) of dreaming do you believe is the best explanation for why we dream​
Semmy [17]

Answer:that your subconscious is trying to tell you something

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3 0
3 years ago
43. A rocket sled accelerates at a rate of 49.0m/s2. Its passenger has a mass of 75.0 kg. (a) Calculate the horizontal component
lord [1]

(a) 3675 N

Assuming that the acceleration of the rocket is in the horizontal direction, we can use Newton's second law to solve this part:

F_x = m a_x

where

F_x is the horizontal component of the force

m is the mass of the passenger

a_x is the horizontal component of the acceleration

Here we have

m = 75.0 kg

a_x = 49.0 m/s^2

Substituting,

F_x=(75.0)(49.0)=3675 N

(b) 3748 N, 11.3 degrees above horizontal

In this part, we also have to take into account the forces acting along the vertical direction. In fact, the seat exerts a reaction force (R) which is equal in magnitude and opposite in direction to the weight of the passenger:

R=mg=(75.0)(9.8)=735 N

where we used

g=9.8 m/s^2 as acceleration of gravity.

So, this is the vertical component of the force exerted by the seat on the passenger:

F_y = 735 N

and therefore the magnitude of the net force is

F=\sqrt{F_x^2+F_y^2}=\sqrt{3675^2+735^2}=3748 N

And the direction is given by

\theta = tan^{-1}(\frac{F_y}{F_x})=tan^{-1}(\frac{735}{3675})=11.3^{\circ}

4 0
3 years ago
An infinite sheet of charge is located in the y-z plane at x = 0 and has uniform charge denisity σ1 = 0.51 μC/m2. Another infini
NNADVOKAT [17]

Answer:

 E_total = 5.8 10⁴ N /C

Explanation:

In this problem they ask to find the electric field at two points, the electric field is a vector magnitude, so we can find the field for each charged shoah and add them vectorally at the point of interest.

To find the electric field of a charged conductive sheet, we can use the Gauss law,

        Ф = E. d S = q_{int} / ε₀

Let us use as a Gaussian surface a small cylinder, with the base parallel to the sheet, the electric field between the sheet and the normal one next to the cylinder has 90º, so its scalar product is zero, the electric field between the sheet and the base has An Angle of 0º, therefore the scalar product is reduced to the algebraic product.

Let's look for the electric field for plate 1

The total flow is the same for each face, as there are two sides of the cylinder

       2E A = q_{int} /ε₀

For the internal load we use the concept of surface density

      σ = q_{int1} / A

      q_{int1} = σ₁ A

Let's replace

       2E A = σ₁ A /ε₀

        E₁ = σ₁ / 2ε₀

For the other plate we have a field with a similar expression, but of negative sign

       E₂ = -σ₂ / 2ε₀

The total field is,

        E_total = σ₁ / 2ε₀ + σ₂ / 2ε₀

       E_total = (σ₁ + σ₂) / 2ε₀

Let us apply this expression to our case, when placing a sheet without electric charge, a charge is induced for each sheet, the plate 1 that has a positive charge the electric field is protruding to the right and the plate 2 that has a negative charge creates a incoming field, to the right, as the two fields have the same address add

           The conductive sheet in the middle pate undergoes an induced load that is created by the other two plates, but because the conductive plate the charges are mobile and are replaced.

       E_total = (0.51 +0.52) 10⁻⁶ / 2 8.85 10⁻¹²

       E_total = 5.8 10⁴ N /C

Note that the field is independent of the distance between the plates

4 0
3 years ago
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